关于C数据类型和常量的问题 [英] Question about C datatype and constant
问题描述
问候！
我正在尝试用C语言，直到遇到一些非常奇怪的东西。
我无法向自己解释下面显示的结果。
Greetings! I was experimenting with C language till I encountered something very strange. I was not able to explain myself the result shown below.
代码：
#include <stdio.h>
int main(void)
{
int num = 4294967295U;
printf("%u\n", num);
return 0;
}
问题：
1。）如你所见，我创建了一个 int ，它可以保存 -2147483648 到
1.) As you see, I created an int which can hold numbers between -2147483648 to 2147483647.
2。）当我将值 4294967295 分配给此变量时，IDE会在此期间向我显示警告消息编译，因为变量溢出。
2.) When I assign the value 4294967295 to this variable, the IDE shows me a warning message during compilation because the variable overflowed.
3。）由于好奇，我在数字后面添加了一个 U（无符号），当我重新编译它时，编译器没有返回任何警告信息。
3.) Due to curiosity I added a U (unsigned) behind the number and when I recompiled it, the compiler did not return any warning message.
4。）我通过将 U（无符号）更改为 L进行了进一步的实验（长）和 LL（很长）。正如预期的那样，警告消息仍然存在于这两个中，但在我将其更改为 UL（无符号长）和 ULL（无符号长长）之后仍然存在。
4.) I did further experiments by changing the U (unsigned) to L (long) and LL (long long). As expected, the warning message still persist for these two but not after I change it to UL (unsigned Long) and ULL (unsigned long long).
5。）为什么会发生这种情况？
5.) Why is this happening?
警告信息:(对于步骤2）
warning #2073: Overflow in converting constant expression from 'long long int' to 'int'.
警告信息:(对于步骤4 LL & L ）
The Warning Message:(For steps 4 LL & L)
warning #2073: Overflow in converting constant expression from 'long long int' to 'long int'.
最后，感谢您阅读我的问题，非常感谢您的教导和建议。
推荐答案
也许，默认情况下，编译器假定您输入的是有符号整数。当你给它 4294967295 时，这个数字不适合4字节整数，所以它使用一个8字节的整数来存储它。然后它必须进行有损转换（ long long ，AKA 8-byte， long ，AKA 4-byte ），所以它会给你一个警告。
Perhaps, by default, the compiler assumes you're typing in signed integers. When you give it 4294967295, that number doesn't fit into a 4-byte integer, so it uses an 8-byte integer to store it, instead. Then it has to do a lossy conversion (long long, AKA 8-byte, to long, AKA 4-byte), so it gives you a warning.
但是，当你输入 4294967295U 时，它知道你想要一个无符号的整数。该数字适合4字节无符号整数，因此它的类型为 long int ，并且不需要有损转换。 （你不会通过从 unsigned long int 到 long int 来丢失数据，只是错误地表示它。）
However, when you type 4294967295U, it knows you want an unsigned integer. That number fits into a 4-byte unsigned integer, so it has type long int, and no lossy conversion is necessary. (You're not losing data by going from unsigned long int to long int, just mis-representing it.)
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